Mech 03 very usefull klepner and kolenkow presentation

7/29/2019 Mech 03 very usefull klepner and kolenkow presentation

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Chapter 3

(Momentum)

Must-do-problems:7, 9, 10, 12, 14, 16,18, 20


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Learning Objectives

Dynamics of system of particles and

conservation of momentum

Center of mass and its motion

Center of mass coordinate

Motion of systems with variable mass


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System of Particles

Consider a system of particles:

o

r1r2

rj

Q. What is the force on the jth particle?

extj

intjj fff

Force on the jth particle due

to all other particles


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Equation of Motion

Q. How one should write the equation of motionfor the system of particles ?

extj

intjjj ffpf

Must be summed for all particles


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ProcessTake each and every particle:

ext

N

int

NN

ext

1

int

11

ffp

.....................

.....................

ffp

j

extj

j

intj

j

j ffp

j=1,2..N

On summing


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Simplification

1) Internal forces are action-reaction forces(Electrostatic, Gravitational.)

jiij ff

0fjint

j

ext

j

extj Ff

2) External forces


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Final Form

ext

ext

j

j

FP

Fp

Where,

Ppdt

dp

j

j

j

j

is the total momentum of the system

(1)


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Bola (Ex. 3.1)

Motion of a BOLA:

Difficulty: To write equation of motion of each as force of

constraint is a function instantaneous positionWay out: Total momentum obeys simple equation

gMgmgmffFP 21ext

2

ext

1ext

..the killer instinctively aims it like a SINGLE MASS


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Snow-capped mechanics!

Motion as good as a single particle

KEY: Where is that SINGLE particle located?


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Way to CM

Lets look at two eq.ext

jj fp

extFP

Unless, known before, the equations are telling you

the same story: Motion of a SINGLE PARTICLE

RMFP ext

We just write,

.and look to define this

new vector quantity


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Way to CM

Recall:j

j

j

j

j rmpP

j

j

jrmRMP

This is true if,

j

j

jrmM

1R


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Center of Mass

o

r1 r2rj

j

j

j

jj

m

rm

R

Position vectorof center of mass

CM

R

N21

NN2211

m......mm

xm.......xmxmX

N21

NN2211

m......mm

ym.......ymymY

N21

NN2211

m......mm

zm.......zmzmZ


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Useful Formula

1. For systems consisting of discrete point masses:

j

jj rmM

1R

2. For continuous systems :

dVrM

1

dArM

1

dmrM

1

R

Also watch out. If or are constants or not. Theyare generally so but need not be (Ref: prob 3.1)


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Example

Find the CM of a semicircular disc of radius a

Q. Does the symmetry of the problem tell you aboutchoosing dA to make your life simple?

dAdAyy

0x

CM

CM

Pure logical guess

y

xa-a

dA


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Solutiony

x

a-a

dA

y

x

a-a

dA=rddr

R0:r;0:

3

a4,0Y,X

CMCMAns:

drrd

drrdsinr

dA

dAyyCM


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Cartesian: Source of error

y

xa-a

dA=dxdy

dxdy

ydxdy

dA

dAyyCM

Settle limits of integration:

???:y;aa:x

22 xa0:y

Source oferror!


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CM CoordinateTo find the motion of a system, coordinate system

is a must and for many body system, coordinate

system fixed at CM is very useful

R

r2

r1

m2

m1R

r2

r1

m2

m1

21

2211

mm

rmrmR

2211 rmrm0R

CM coordinate

system

x

y

z

y

x

z


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Advantage CM Coordinate

22112211 rmrmrmrm0R

If the motion of one particle is known, the

motion of the other particle follows directly

1

.constV0FFPrmRM CMextextj

jj

Motion of CM independent of internal forces !

With no ext force, the CM system is inertial !

VCM once known, it is forever!

2


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Ex 3.8

Two identical blocks (a and b) of mass m each,

connected by a spring of spring constant kand

moving on a frictionless track. Block a is given an

initial velocity ofv0 to the right. Determine the

subsequent motion of the two blocks. Initially they

were at rest.

va=v0

O

b

rb ra

a

The Push Me Pull You problem

(You are advised to work out this before jumping to 3.7.)


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Solution

Essence of the solution:The answer (motion of the two blocks) must come in terms

of given quantities, all in the frame where you do the expt.

Q1. Can CM be a good reference frame ?

Q2. If so, how does it ease out things?

The KEY:

As system slides freely (frictionless), so the CMcoordinate system safely defines an inertial frame

Define CM and rewrite coordinates with its respect


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Search for CM

baba rr21m2 mrmrR

(halfway thro a and b)

CM coordinatefor a and b:

ababb

baaa

rrr2

1Rrr

rr

2

1Rrr

b

rb ra

a

R O

br ar

CM coordinatex


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Solution

Instantaneous lengthof the string: lrrlrr baba

The equations of motion in CM system:

lrrkrm

lrrkrm

bab

baa

Unstretched lengthof the string: l


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Solution

Subtracting one from the other: lrrk2rmrm baba

tcosBtsinAtu Solutionm

k2

0ku2um u: A variable that tells us the departure of the spring

from its equilibrium length


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Constants

1. Spring unstretched at t=0 u(0) = 0 B = 0

lrrlrru baba 2. As

0ba v0cosA0v0v0u 0v

A

The Final solution:

tsinv

tu 0

tcosBtsinAtu


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Further

Since,

baba

0ba

vvrr

tcosvuvv

AND,

tcosv21vv 0ba

This follows,


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More Information

Laboratory and CM velocities:

?R

;vRv;vRv bbaa

No change from the value at t=0!

0ba v21

0v0v2

1R

ba rr2

1R


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Finally,

tcos12

vv

tcos1

2

vv

0

b

0

a

The masses move to theright on the average butthey alternatively come to

rest in a push me-pull

you fashion.

Clubbing equations

0bbaa v2

1

R;vRv;vRv

tcosv21vv 0ba


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Known Stuff

Examine some of apparently known scenario :

4. A leaky wagon

1. Conveyer belt in airport

2. Rocket Motion

3. A snowball gaining in size while rollingdown a snowcapped hill

5. An evaporating raindrop

What is common in all?


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The Problem

A spacecraft moves through the space with constant

velocity v. It encounters a stream of dust particles that

embed themselves in it at rate dm/dt . The dust hasvelocity u just before it hits. At time t the mass of the

space-craft is M(t). Find the ext. force F necessary to

keep the spacecraft moving uniformly.

Example: 3.11

Q. Can we apply blindly to analyze forceson system with variable masses?

vmdtd

F


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Concept Behind

vM(t)

System Boundary :

M + m

vM+m

m

F

u

System Boundary :

M + m

m to be added in time t


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Solution

Initial Momentum: umvtMtP

Final Momentum: vmvtMtmtMttP

Then, it follows,

dt

dmuvF

dt

Pd


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Notable Point

When differentials are involved, for system withvariable masses, the known equation involving

velocity and momentum for single mass is not

always appropriate Be watchful !

Erroneous ! (Ref: previous example)

dt

dmuvdt

dmvdt

dmvdt

vdmvmdt

dF


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Force Eq. (Revisit)

dt

dmuvF

dt

Pd

u: Initial velocity of the mass being added

v: Velocity of the ship where mass is being

added

dm/dt: Rate at which the mass being added


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F

Ex 3.12

v

Sand falls from a stationed

hopper at a rate dm/dt on to

a car which is moving withuniform vel v. How much is

force reqd to keep the car

moving at v?


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Solution

The Key:It is a variable mass problem and the initial velocity

of the sand is zero as hopper is at rest u = 0

dtdmv

dtdmuvF

It quickly follows:


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Rockets

Mass Continually Leaving the system

Rocket in free space. What is most striking about it?

With no external agent to push on or be pushedby, how does an object get itself moving?

Carefully examine the rocket eq.as a variable mass problem


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Rocket Motion

M

v

m M

v+v

v+v+u

m

At time t At time t+t

1. Look at the problem as a system

marked by the dotted boundary

2. u is the exhaust vel wrt rocket

Key :


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Rocket Motion Contd.

vmMtP

uvvmvvMttP

M

v

m M

v+v

v+v+u

m

At time t At time t+t

*u taken positive in the direction ofv

k d


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Hence,dtdmu

dtvdM

dtPd

dt

dMu

dt

vdMFext

Fundamental Rocket Equation

But, mass comes from rocket dt

dM

dt

dm


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Recall our claim: That rocket motion tells ushow without a push or pull, an object propels

We must see what happens at,

0Fext

Rocket in free space

(Ex. 3.14)

(Ex. 3.15: Rocket motion in gravitational field

is left to you as home assignment)


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0dt

dMudt

vdMF

MduvM

Just like Newtons 2nd law for a particle, except that

the product on the right plays the role of the force

ThrustMdu


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Mathematical Solution

dtdM

Mu

dtvd

f

0

0

f

0fM

Mlnu

M

Mlnuvv

After integration (taking exhaust vel const.),

v0= initial velocity of rocket

M0= initial mass of the rocket

l P


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Two Critical Points

f

0

0fMMlnuvv

f

0

fMMlnuv for v0=0

2. This equation puts a significant restrictionon the maximum speed of the rocket.

1. The final velocity is independent ofhow

mass is released.

Why and How do you improve on it?


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Velocity Restriction

Rewrite rocket eqn. :

f

00fM

Mlnuvv

If the original mass is 90% fuel, the ratio is 10

vf-v0 CANNOT be more than 2.3 times u

Then,

Q. What should the rocket engineers do?

Mech 03 very usefull klepner and kolenkow presentation

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